3k-2 ,4k-6, and k+2 are three consecutive terms of AP, then find the value of k .Also if the sum of first terms of an AP is 63 and sum of its next 7 terms is 161. Find 28th term of AP.
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It is given that k+2,4k−6,3k−2 are the consecutive terms of A.P, therefore, by arithmetic mean property we have, first term+third term is equal to twice of second term that is:
(k+2)+(3k−2)=2(4k−6)
⇒4k=8k−12
⇒4k−8k=−12
⇒−4k=−12
⇒4k=12
⇒k= 12/4=3
Now,
Sum of n terms of an A.P. = n/2 (2 a + (n-1)d),
where a is the 1st term of A.P and d is the common difference
So,sum of 1st 7 terms=7/2 (2 a + 6d)=63
2a+6d = 18 —(equation 1)
Sum of next 7 terms=
sum of 1st 14 terms- sum of 1st 7 terms =
14/2 (2 a + 13d) - 7/2 (2 a + 6d)
So, 161 = 14a +91d - 63
224 =14 a +91d—(equation 2)
(Eq. 1 )*7 - (eq. 2)
14 a + 42d- 14a - 91d =126-224
49d = 98
d = 2,a = 3
Nth term = a +(n-1)d
28th term = 3 + 27×2 = 57
(k+2)+(3k−2)=2(4k−6)
⇒4k=8k−12
⇒4k−8k=−12
⇒−4k=−12
⇒4k=12
⇒k= 12/4=3
Now,
Sum of n terms of an A.P. = n/2 (2 a + (n-1)d),
where a is the 1st term of A.P and d is the common difference
So,sum of 1st 7 terms=7/2 (2 a + 6d)=63
2a+6d = 18 —(equation 1)
Sum of next 7 terms=
sum of 1st 14 terms- sum of 1st 7 terms =
14/2 (2 a + 13d) - 7/2 (2 a + 6d)
So, 161 = 14a +91d - 63
224 =14 a +91d—(equation 2)
(Eq. 1 )*7 - (eq. 2)
14 a + 42d- 14a - 91d =126-224
49d = 98
d = 2,a = 3
Nth term = a +(n-1)d
28th term = 3 + 27×2 = 57
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